Problem: Find $\lim_{x\to 1}\dfrac{x^4+2x^3-3x^2}{x^3-x^2}$. Choose 1 answer: Choose 1 answer: (Choice A) A $4$ (Choice B) B $2$ (Choice C) C $0$ (Choice D) D The limit doesn't exist
Solution: Substituting $x=1$ into $\dfrac{x^4+2x^3-3x^2}{x^3-x^2}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression on our hands, let's try to simplify it. $\dfrac{x^4+2x^3-3x^2}{x^3-x^2}$ can be simplified as $x+3$, for $x\neq 0,1$. This means that the two expressions have the same value for all $x$ -values (in their domains) except for $0$ and $1$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{x^4+2x^3-3x^2}{x^3-x^2}=x+3$ for all $x$ -values in the interval $(0,2)$ except for $x=1$. Therefore, $\lim_{x\to 1}\dfrac{x^4+2x^3-3x^2}{x^3-x^2}=\lim_{x\to 1}(x+3)=4$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to 1}\dfrac{x^4+2x^3-3x^2}{x^3-x^2}=4$.